General technique to compute the covering $S^1\vee S^1$ asociated to certain normal subgroup of $F_2$

Question

In p.80 of Hatcher's book we can find the following exercise:

Exercise 1.3.12. Let $a$ and $b$ be the generators of $\pi_1(S^1\vee S^1)$ corresponding to the two $S^1$ summands. Draw a picture of the covering space of $S^1\vee S^1$ corresponding to the normal subgroup generated by $a^2$, $b^2$ and $(ab)^4$, and prove that this covering space is indeed the correct one.

By Van Kampen's theorem $\pi_1(S^1\vee S^1)$ is the free group $F_2=\langle a,b\rangle$ generated by $a$ and $b$. In order to solve the exercise we have to remember the following facts:

1. Every properly discontinuous group action $G\to \text{Hom}(X)$ over a topological space $X$ induces a normal covering $p:X\to X/G$ via the quotient map from $X$ to the orbit space $X/G$ (see Proposition 1.40 (a) of Hatcher's book)
2. A presentation group $G=\langle g_\alpha\, |\, r_\beta\rangle$ acts over its Cayley graphs by a properly discontinuous action (see p.77 of Hatcher's book)

In order compute the covering space associated to a normal subgroup $H$ of $F_2$, I am going to show that it is sufficient to compute the Cayley graph of $F_2/H$:

Claim. Let $H$ be a normal subgroup of the free group $F_2$ with two generators. And $p$ the normal covering induced by the natural action of $G=F_2/H$ over the Cayley Graph $K$ of $G$. Then $p$ is the covering map corresponding to the subgroup $H$, that is: $$p_\star\big(\pi_1(K)\big)\cong H.$$

Proof. Let $K$ be the Cayley graph of the quotient $G=F_2/K$ and $p:K\to K/G$ the normal covering described above for the natural action of $G$ over $K$. By the construction of the Cayley graph, it follows that the orbit space must be trivially homeomorphic to the wedge sum $S^1\vee S^1$. So we have a normal covering: $$p:K\to S^1\vee S^1$$ We have to show that this covering is indeed the correct one. For do so we know that

• Since $K$ is path connected, the transformation group $\mathcal G$ is isomorphic to $G=F_2/H$ (see Proposition 1.40 (b))
• $p$ is normal so, the transformation group $\mathcal G$ is isomorphic to $F_2/p_\star(\pi_1(K))$ (see Proposition 1.39) Thus: $$F_2/H\quad \text{is isomorphic to}\quad F_2/p_\star(\pi_1(K))$$ which implies $H\cong p_\star(\pi_1(K))$.

$\square$

• Is correct my reasoning?
• Is there an easier way to compute in general that covering map?

Thanks in advance.